Two pairs of concentric circular coils are placed in the same...

The induced emf in the inner coil is given by $E = M \dfrac{dI}{dt}$, where $M$ is the mutual inductance between the outer and inner coils.

The magnetic field at the centre due to the outer coil is $B = \dfrac{\mu_0 I}{2R}$.
The flux linked with the inner coil is $\Phi = B \times \pi r^2 = \left(\dfrac{\mu_0 I}{2R}\right) \pi r^2$.
Thus, the mutual inductance is $M = \dfrac{\Phi}{I} = \dfrac{\mu_0 \pi r^2}{2R}$.

For Pair A, the mutual inductance is:
$M_A = \dfrac{\mu_0 \pi r^2}{2R}$

For Pair B, the inner radius is $2r$ and the outer radius is $2R$. The mutual inductance is:
$M_B = \dfrac{\mu_0 \pi (2r)^2}{2(2R)} = \dfrac{\mu_0 \pi (4r^2)}{4R} = \dfrac{\mu_0 \pi r^2}{R}$

Comparing $M_A$ and $M_B$, we see that $M_B = 2 M_A$.

Since the rate of change of current $\dfrac{dI}{dt}$ is the same for both pairs, the ratio of the induced emfs is simply the ratio of their mutual inductances:
$\dfrac{E_A}{E_B} = \dfrac{M_A}{M_B} = \dfrac{M_A}{2M_A} = \dfrac{1}{2}$.

Answer: $1/2$